Lemma 5 Let f be n-times differentiable on . If , then for every λ, , we have
(3.1)
Proof Let p, , be the integer assigned to the function f as in Lemma 1. Because of , we always have . Furthermore, let be assigned to f by Lemma 1. Then, by using the Taylor formula on time scales, for every , we obtain
(3.2)
By using Theorem 2 and (3.2), we have
Since f is n-times differentiable on and with , we have with n and f substituted by and , respectively
Also, for every , s with and , we have
This is obvious for and, when , it can be derived by applying the Taylor formula. Thus, for all , we get
and therefore the proof of the lemma can be immediately completed. □
The result of Lemma 5 is an extension of studies in [4] and [5]. In order that the reader sees how the results in [4] (1.8.14) and [5] (Lemma 2) follow from (3.1), it is at this point only necessary to know that in the case , we have , and
then we get the inequality in [4]
and in the case , we have and
then we get the inequality in [5]
For the cases and , some sufficient criterias for oscillatory behaviour of the solutions of the equation (1.1) were obtained by Bolat and Akın in [6] and [7], respectively. Furthermore, there might be other time scales that we cannot appreciate at this time due to our current lack of ‘real-world’ examples.
Theorem 3
Assume that
n
is odd and
(C1) ,
(C2) .
Then every bounded solution of equation (1.1) is either oscillatory or tends to zero as .
Proof Assume that equation (1.1) has a bounded non-oscillatory solution . Without loss of generality, assume that is eventually positive (the proof is similar when is eventually negative). That is, , and for and . Assume further that does not tend to zero as . By (1.1), (1.2), we have for
(3.3)
It follows that () is strictly monotone and eventually of constant sign. Since is an oscillatory function, there exists a such that if , then . Since is bounded, by virtue of (C1) and (1.2), there is a such that is also bounded for . Because n is odd and is bounded, by Lemma 1, when (otherwise is not bounded), there exists such that for we have , .
In particular, since for , is decreasing. Since is bounded, we write (). Assume that . Let . Then there exists a constant and a such that for . Since is bounded, by (C1). Therefore, there exists a constant and a such that for . So that we can find a with such that for . From (3.3) we have
(3.4)
for . If we multiply (3.4) by and integrate it from to t, we obtain
(3.5)
where
and
Since for and , we have for . From (3.5) we have
By (C2) we obtain
as . This is a contradiction. So, is impossible. Therefore, is the only possible case. That is, . Since is bounded, by (C1) we obtain
from (1.2).
Now let us consider the case of for . By (1.1) and (1.2),
for . That is, . It follows that () is strictly monotone and eventually of constant sign. Since is an oscillatory function, there exists a such that if , then . Since is bounded, by (C1) and (1.2), there is a such that is also bounded for . Assume that . Then . Therefore, and for . Hence, we observe that is bounded. Since n is odd, by Lemma 1, there exists a and (otherwise is not bounded) such that , and . That is, , and . In particular, for we have . Therefore, is increasing. So, we can assume that (). As in the proof of , we may prove that . As for the rest, it is similar to the case of . That is, . This contradicts our assumption. Hence the proof is completed. □
Theorem 4 Assume that n is even and (C 1) holds. If the following condition is satisfied:
(C3) There is a function such that . Moreover,
and
for and . Then every bounded solution of equation (1.1) is oscillatory.
Proof Assume that equation (1.1) has a bounded non-oscillatory solution . Without loss of generality, assume that is eventually positive (the proof is similar when is eventually negative). That is, , and for . By (1.1) and (1.2), we have (3.3) for . Then . It follows that () is strictly monotone and eventually of constant sign. Since is an oscillatory function, there exists a such that for , we have . Since is bounded, by (C1) and (1.2), there is a , such that is also bounded for . Because n is even, by Lemma 1 when (otherwise is not bounded), there exists such that for we have
(3.6)
In particular, since for , is increasing. Since is bounded,
by (C1). Let ; i.e., there exists a such that by (1.2)
for . We may find a such that for and ,
(3.7)
From (3.3), (3.7) and the properties of f, we have
(3.8)
for . Since is bounded and increasing, (). By the continuity of f, we have
Then there is a such that for , , we have
(3.9)
By (3.8), (3.9),
(3.10)
Set
(3.11)
We know from (3.6) that there is a such that for a sufficiently large , . Therefore, Δ-derivating (3.11) we obtain
(3.12)
(3.13)
We know from (3.6) that there is a such that and for an even n. Since is increasing for . Therefore, by Lemma 5,
(3.14)
Then by Δ-derivating (3.14) and using , we get
by Lemma 2
Since , we obtain
(3.15)
Hence by (3.10), (3.14) and (3.15), we have
and then
(3.16)
for . If we multiply (3.16) by and integrate it from to t, we obtain by Theorem 2
Therefore by (C3),
This is a contradiction.
Now let us consider the case of for . By (1.1) and (1.2), we have
for . That is, . It follows that () is strictly monotone and eventually of constant sign. Since is an oscillatory function, there exists a such that for . Since is bounded, by (C1) and (1.2), there is a such that is also bounded for . Assume that . Then . Therefore, and for . Hence, we observe that is bounded. Since n is odd, by Lemma 1, there exists a and (otherwise is not bounded) such that , and . That is, , and . In particular, for we have . Therefore, is increasing. For the rest of the proof, we can proceed the proof similarly to the case of . Hence, the proof is completed. □